June 2023 Paper 1 Q10

OCR ACurrent spec9 marksNumerical Methods

Graph of f(x) for x from about -0.3 to 4: a branch with a small maximum near the origin that drops steeply below the x-axis, crossing it just after x = 0; and a branch that comes down steeply from above, just after x = 0.5, levels off just above the x-axis and rises slowly towards x = 4

The diagram shows part of the curve \(\mathrm{f}(x) = \dfrac{\mathrm{e}^x}{4x^2 - 1} + 2\). The equation \(\mathrm{f}(x) = 0\) has a positive root \(\alpha\) close to \(x = 0.3\).

(a) Explain why using the sign change method with \(x = 0\) and \(x = 1\) will fail to locate \(\alpha\). [1]

(b) Show that the equation \(\mathrm{f}(x) = 0\) can be written as \(x = \dfrac{1}{4}\sqrt{\left(4 - 2\mathrm{e}^x\right)}\). [2]

(c) Use the iterative formula \(x_{n+1} = \dfrac{1}{4}\sqrt{\left(4 - 2\mathrm{e}^{x_n}\right)}\) with a starting value of \(x_1 = 0.3\) to find the value of \(\alpha\) correct to 4 significant figures, showing the result of each iteration. [3]

(d) An alternative iterative formula is \(x_{n+1} = \mathrm{F}(x_n)\), where \(\mathrm{F}(x_n) = \ln\left(2 - 8x_n^{\,2}\right)\).
By considering \(\mathrm{F}'(0.3)\) explain why this iterative formula will not find \(\alpha\). [3]

Mark scheme (a)

Scheme	Marks	AO
Both \(\mathrm{f}(0)\) and \(\mathrm{f}(1)\) are positive so no sign change will be seen	B1	2.3
	[1]

Notes

B1: Identify both \(y\)-values being positive and state ‘no sign change’ or equiv
Could also evaluate \(\mathrm{f}(0)\) as 1 and \(\mathrm{f}(1)\) as 2.9 (or better), and refer to no sign change – these are both positive so no need to include \(\gt 0\)
Could also refer to the asymptote / discontinuity within this range (\(x = 0\) to \(x = 1\))
Also allow ‘graph is not continuous in this interval’
B0 for no reference to interval
Could say that the two points chosen are not on the same part of the curve

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{\mathrm{e}^x}{4x^2 - 1} = -2\) \(\mathrm{e}^x = -8x^2 + 2\) \(8x^2 = 2 - \mathrm{e}^x\)	M1	1.1
\(16x^2 = 4 - 2\mathrm{e}^x\) \(4x = \sqrt{4 - 2\mathrm{e}^x}\) \(x = \dfrac{1}{4}\sqrt{\left(4 - 2\mathrm{e}^x\right)}\) A.G.	A1	1.1
	[2]

Notes

M1: Attempt rearrangement, as far as \(kx^2 = \ldots\)
Allow sign error(s) only

A1: Obtain given answer convincingly
If \(x = \sqrt{\dfrac{1}{4} - \dfrac{1}{8}\mathrm{e}^x}\) then an additional line of working needed before given answer (eg show common denominator of 16)

Mark scheme (c)

Scheme	Marks	AO
\(x_2 = 0.285074813\ldots\)	B1	1.1
0.28943, 0.28817, 0.28853, 0.28843, 0.28846, 0.28845...	M1	1.1
\(\alpha = 0.2885\)	A1	1.1
	[3]

Notes

B1: Correct first iterate (at least 4sf)
State 0.2851 or better

M1: Correct iterative process (at least 3 more values)
Allow M1 for 3sf – expect 0.289, 0.288 and then 0.288 or 0.289 depending whether truncating or rounding

A1: Correct root, given to 4sf, following 2 iterates that agree to 4sf
ie at least 7 iterations needed, given to at least 4sf
A0 for eg \(x_8 = 0.2885\) (implies 8^th iterate and not root)
Process self corrects so B0M1A1 possible; or B1M1A1 if error in term other than \(x_2\)

Mark scheme (d)

Scheme	Marks	AO
\(\mathrm{F}'(x) = \dfrac{-16x}{2 - 8x^2}\)	M1	1.1a
\(\mathrm{F}'(0.3) = -3.75\)	M1	1.1
For convergence \(\|\mathrm{F}'(\alpha)\| \lt 1\), but \(-3.75 \lt -1\), so iteration will not find root	A1	2.5
	[3]

Notes

M1: Attempt differentiation using the chain rule
Obtain derivative of form \(\dfrac{kx}{2 - 8x^2}\)
Condone subscripts still present in derivative

M1: Attempt \(\mathrm{F}'(0.3)\) – not dependent on previous M1, but must follow some attempt at differentiation
M1 can be implied by correct \(-3.75\) (from correct derivative), but explicit substitution must be seen if \(\mathrm{F}'(x)\) is incorrect
Must come from differentiating \(\mathrm{F}(x)\) and not a different function

A1: Correct reasoning, following correct \(\mathrm{F}'(0.3)\)
Allow \(\mathrm{F}'(\alpha) \lt -1\), hence will not converge
Condone \(\mathrm{F}'(x)\) not \(\mathrm{F}'(\alpha)\)
No credit for just testing the given iterative formula